Question: $\int^{1/3}_{-1/3}\dfrac{1}{1+9x^2}\,dx\, = $
Solution: Strategy Let's first find the indefinite integral $\int\dfrac{1}{1+9x^2}\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral First, notice that we can rewrite the integral as $\int\dfrac{1}{1+(3x)^2}\,dx\, $. If we let $ {u=3x}$, then ${du=3 \, dx}$ and ${ dx=\dfrac{1}{3}\, du}$. So we have: $\begin{aligned}\int\dfrac{1}{1+9x^2}\,dx\, &=\int\dfrac{1}{1+({3x})^2}\,{dx}\,\\\\\\\\ &=\int\dfrac{1}{1+ u^2}\,\cdot {\dfrac13\, du}\,\\\\\\\\ &=\dfrac13\int\dfrac{1}{1+u^2}\,du\\\\\\\\ &=\dfrac{1}{3}\tan^{-1}(u)+C\\\\\\\\ &=\dfrac{1}{3}\tan^{-1}(3x)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{1/3}_{-1/3}\dfrac{1}{1+9x^2}\,dx\, &= \dfrac{1}{3}\tan^{-1}(3x)\Bigg|^{1/3}_{-1/3}\\\\\\\\ &=\dfrac{1}{3}\left(\tan^{-1}(1)-\tan^{-1}(-1)\right)\\\\\\\\ &=\dfrac13\left(\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}\right)\right)\\\\\\ &=\dfrac13\left(\dfrac{\pi}{2}\right)\\\\\\\\ &=\dfrac{\pi}{6}\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{1/3}_{-1/3}\dfrac{1}{1+9x^2}\,dx\, = \dfrac{\pi}{6}$